# Mathimatical Induction

1. Aug 4, 2009

### logicaljoe

Let P(n) be the formula

1^2 + 2^2 + ... + n^2 = n(n +1)(2n +1)/6

a. Write P, is it true?(1)

1^2 + 2^2 + ... + 1^2 = 1(1 +1)(2(1) +1)/6 =

1^2 + 2^2 + ... + 1^2 = 1(1 +1)(2 +1)/6

LHS = 1^2 = 1

RHS = 2 . 3/6 = 1 True

b. Write P(k)

Basis Step

n = k

1^2 + 2^2 + ... + k^2 = k(k +1)(2k +1)/6

c. Write P(k+1)

Inductive step

n = k then n = k + 1

1^2 + 2^2 + ... + (k+1)^2 = (k+1)((k+1) +1)(2(k+1)+1)/6

Now this is where I get confused with induction writing the proof of the inductive step.

d. use mathematical induction to prove k then k+1, n>_1

1^2 + 2^2 + ... + (k+1)^2 = (k+1)((k+1) +1)(2(k+1)+1)/6

LHS = (k+1)^2
= k+1
= k
= 1

RHS = (k+3)+1(2+1)/6 combining like terms
= (k+3)+(2+1)/6
= 3k+3/6
= 6k/6
= k
= 1

Is this the kind of working that is valid? it seems right to me.

2. Aug 4, 2009

### jgens

This seems like a homework question (it's a problem in Spivak's Calculus). If it is, for future reference, these threads should be placed in the homework help sections: https://www.physicsforums.com/forumdisplay.php?f=155

This seems overly complicated and doesn't appear quite right (you seem to be confused about the inductive step). For the inductive step, assuming the P(k) is true, we need to demonstrate that this implies that P(k + 1) is true. I can help you complete the proof from what you've got.

Proof: Let $$P(x)$$ be the statement that, for some natural number $$x$$,

$$1^2 + 2^2 + . . . + (x - 1)^2 + x^2 = \frac{x(x+1)(2x + 1)}{6}$$

Clearly $$P(1)$$ is true since,

$$1^2 = \frac{1(2)(3)}{6} = \frac{6}{6} = 1$$

Now, suppose that $$P(x)$$ is true for some arbitrary natural number $$k$$. To complete the proof, we need only show that $$P(k)$$ implies $$P(k + 1)$$. Therefore,

$$1^2 + 2^2 + . . . + (k - 1)^2 + k^2 + (k + 1)^2 = \frac{k(k+1)(2k + 1)}{6}+ (k + 1)^2$$

Try and complete the proof from here.